Question

1. A solution contains 1 billion 3p atoms (half life of S2p is 14.3 days) and it p-decays with a maximum energy to 32s. (a) what is the activity, in becquerels, of this solution? (b) how many atoms of phosphorus and of sulphur are in this solution in one week? (c) how long will it take for only 100 P atoms to remain in this solution?

Answer 1

Initial number of atoms $N_0=1\,\text{billion}$

Half life of atoms $t_{1/2}=14.3\,days$

$t_{1/2}=\frac{\ln 2}{\lambda}$

Number of atoms at time $t$, $N=N_0e^{-\lambda t}$

Activity of atoms $A=\left|\frac{dN}{dt}\right|=\lambda N_0e^{-\lambda t}=A_0e^{-\lambda t}$

(a)

Initial activity $A_0=\lambda N_0=\left(\frac{\ln2}{14.3*86400} \right )*10^{9}=561\,dps=561\,Bq$

(b)

Number of Phosphorus atoms at time $t=7\,days$

$N=N_0e^{-\lambda t}=(10^9)e^{-\frac{\ln2}{14.3}*7}=0.712*10^{9}=0.712\,\text{billion}$

Number of Sulphur atoms $N_0-N=0.288\,\text{billion}$

(c)

Number of Phosphorus atoms $N=100$ at time $t$ days

$N=N_0e^{-\lambda t}$

$100=(10^9)e^{-\lambda t}=(10^9)e^{-\frac{\ln2}{14.3}*t}$

$e^{\lambda t}=10^{7}$

$\lambda t=\ln\left(10^7 \right )$

$\frac{\ln 2}{14.3} t=\ln\left(10^7 \right )=16.11809565$

Time (in days) $t=\frac{16.11809565*14.3}{\ln 2}=332.5\,\text{days}$